Joint Probability Distributions – Examples Example 1 A certain ...

Joint Probability Distributions – ExamplesExample 1 A certain market has both an express checkout line and a superexpress checkout line.Let X denote the number of customers in line at the express checkout line at a particular timeof day and let Y denote the number of customers in line at the superexpress checkout at thesame time. Suppose the joint pmf of X and Y is as given in the following table:X / Y 0 1 2 30 0.08 0.07 0.04 0.001 0.06 0.15 0.05 0.042 0.05 0.04 0.10 0.063 0.00 0.03 0.04 0.074 0.00 0.01 0.05 0.06a. What is P (X = 1; Y = 1), that is, the probability that there is exactly one customer in eachline?Solution P (X = 1; Y = 1) = p(1; 1) = 0:15:b. What is P (X = Y ), that is, the probability that the number of customers in the two lines areidentical?Solution P (X = Y ) = p(0; 0) + p(1; 1) + p(2; 2) + p(3; 3) = 0:08 + 0:15 + 0:10 + 0:07 = 0:40:c. Let A denote the event that there are at least two more customers in one line than in the otherline. Express A in terms of X and Y and calculate the probability of this event.Solution A = f(x; y) : x ¸ y + 2g [ f(x; y) : y ¸ x + 2gP (A) = p(2; 0) + p(3; 0) + p(4; 0) + p(3; 1) + p(4; 1) + p(4; 2) + p(0; 2) + p(0; 3) + p(1; 3) = 0:22:d. What is the probability that the total number of customers in the two lines is exactly four? Atleast four?Solution The desired probabilities are:P (X + Y = 4) = p(1; 3) + p(2; 2) + p(3; 1) + p(4; 0) = 0:17P (X + Y ¸ 4) = P (X + Y = 4) + p(4; 1) + p(4; 2) + p(4; 3) + p(3; 2) + p(3; 3) + p(2; 3) = 0:46:e. Determine the marginal pmf of X and then calculate the expected number of customers in lineat the express checkout.Solution p X (n) = P (X = n) = p(n; 0) + p(n; 1) + p(n; 2) + p(n; 3)E(X) =x 0 1 2 3 4p X (x) 0.19 0.30 0.25 0.14 0.124Xx ¢ p X (x) = 0(0:19) + 1(0:30) + 2(0:25) + 3(0:14) + 4(0:12) = 1:7:x=01:

f. Determine the marginal pmf of Y .Solution p Y (n) = P (Y = n) = p(0; n) + p(1; n) + p(2; n) + p(3; n) + p(4; n)y 0 1 2 3p Y (y) 0.19 0.30 0.28 0.23g. By inspection of the probabilities P (X = 4), P (Y = 0), and P (X = 4; Y = 0), are X and Yindependent random variables? Explain.Solution P (X = 4) = 0:12, P (Y = 0) = 0:19, and P (X = 4; Y = 0) = 0. Hence there exists anordered pair (x; y) such that p(x; y) 6= p X (x)¢p Y (y) and therefore X and Y are not independentrandom variables.Example 2 An instructor has given a short quiz consisting of two parts. For a randomly selectedstudent, let X be the number of points earned on the …rst part and Y be the number of pointsearned on the second part. Suppose the joint pmf of X and Y is given in the following table:X / Y 0 5 10 150 0.02 0.06 0.02 0.105 0.04 0.15 0.20 0.1010 0.01 0.15 0.14 0.01a. If the score recorded in the grade book is the total number of points earned on the two parts,what is the expected recorded score E(X + Y )?Solution The expected recorded score is:E(X + Y ) = X X(x + y) ¢ p(x; y)x y= (0 + 0)(0:02) + (0 + 5)(0:06) + : : : + (10 + 15)(0:01) = 14:10:Note that E(X + Y ) = E(X) + E(Y ).b. If the maximum of the two scores is recorded, what is the expected recorded score?Solution The expected recorded score is:E [max(X; Y )] = X Xmax(x; y) ¢ p(x; y)x y= (0)(0:02) + (5)(0:06) + : : : + (15)(0:01) = 9:60:c. Compute the covariance for X and Y .Solution The covariance for X and Y is computed as follows:E(X) = X x x ¢ p X(x) = 5:55E(Y ) = X y ¢ p yY (y) = 8:55E(XY ) = X Xxy ¢ p(x; y) = (0)(0:02) + : : : + (150)(0:01) = 44:25x yCov(X; Y ) = E(XY ) ¡ E(X)E(Y ) = 44:25 ¡ (5:55)(8:55) = ¡3:2025:2:

d. Compute the correlation for X and Y .Solution The correlation for X and Y is computed as follows:V (X) = X x (x ¡ ¹ X) 2 ¢ p X (x) = 12:45V (Y ) = X y (y ¡ ¹ Y ) 2 ¢ p Y (y) = 19:15Corr(X; Y ) = ½ XY =¡3:2025q(12:45)(19:15) = ¡0:2074:Example 3 Two components of a minicomputer have the following joint pdf for their useful lifetimesX and Y :( )x exp [¡x(1 + y)] x ¸ 0 and y ¸ 0f(x; y) =:0 otherwisea. What is the probability that the lifetime X for the …rst component exceeds 3?Solution The desired probability is:P (X > 3) =Z 1 Z 130x exp [¡x(1 + y)] dy dx =Z 13exp (¡x) dx = exp(¡3) = 0:0499b. What are the marginal pdf’s of X and Y ? Are the two lifetimes independent? Explain.Solution The marginal pdf’s of X and Y are:f X (x) = R 10 x exp [¡x(1 + y)] dy = exp(¡x) for x ¸ 0f Y (y) = R 10 x exp [¡x(1 + y)] dx =1(1 + y) 2 for y ¸ 0:It is now clear that f(x; y) is not the product of the marginal pdf’s, so the two randomvariables are not independent.c. What is the probability that the lifetime of at least one component exceeds 3?Solution P (at least one component lifetime exceeds 3) = 1 ¡ P (X · 3 and Y · 3)= 1 ¡= 1 4Z 3 Z 300x exp [¡x(1 + y)] dy dx =+ exp(¡3) ¡exp(¡12)4= 0:2998Z 30¡ exp(¡4x) + exp(¡x) dxExample 4 Let X and Y have the joint probability density function given by:f(x; y) =(k(1 ¡ y) 0 · x · y · 10 otherwisea. Find the value of k that makes this a probability density function.):3

Solution For f(x; y) to be a valid joint pdf, R 1 R 1¡1 ¡1 f(x; y) dx dy = 1. HenceZ 1Z yb. Find the marginal pdf’s for X and Y .Solution The marginal pdf’s for X and Y are:f X (x) =0Z 1x0k(1 ¡ y) dx dy = k 6 ) k = 6:6(1 ¡ y) dy = 3x 2 ¡ 6x + 3 = 3 (x ¡ 1) 2c. Are X and Y independent?f Y (y) =Z y06(1 ¡ y) dx = 6y(1 ¡ y):Solution X and Y are not independent since f(x; y) 6= f X (x) ¢ f Y (y).d. Find E(X), E(Y ), V (X), V (Y ), Cov(X; Y ), and Corr(X; Y ).Solution The solution is:V (X) =V (Y ) =E(X) =E(Y ) =Z 10Z 1Z 10Z 10µx ¡ 1 40Z 1 Z yx ¢ 3(x ¡ 1) 2 dx = 1 4y ¢ 6y(1 ¡ y) dy = 1 2 2¢ 3(x ¡ 1) 2 dx = 3 80µy ¡ 1 2¢ 6y(1 ¡ y) dy = 1 220E(XY ) = xy ¢ 6(1 ¡ y) dx dy = 30 020Cov(X; Y ) = E(XY ) ¡ E(X)E(Y ) = 3 µ µ 1 120 4 ¡ 2Corr(X; Y ) = qCov(X; Y )1V (X)V (Y ) = r ³38040´ ³120´ == 140p33 = 0:5774:4